the parallelogram GM equal to the triangle DBC, having GHM BOOK I for one of its angles and because the straight line GH meets the parallels KM, FG, the alternate angles MHG, HGF are equal: add to each of these the angle HGL; therefore the c 29. 1. angles MHG, HGL are equal to the angles HGF, HGL: But MHG, HGL are equal to two right angles; therefore alfo HGF, HGL are equal to two right angles: and because the two ftraight lines FG, GL, on oppofite fides of GH, make the adjacent angles HGF, HGL equal to two right angles, FG is in the fame ftraight line d with GL: And because KF is parallel to HG, and HG to ML; KF is parallel to ML: And KM, FL are parallels; wherefore KFLM e the triangle D C E F B K H M is a parallelogram; and d 14. I. e 30. I. COR. From this it is manifeft, how to a given ftraight line to apply a parallelogram, which shall have an angle equal to a given rectilineal angle, and fhall be equal to a given rectilineal figure, viz. by applying to the given ftraight line, a parallelogram b 44. 1. equal to the first triangle ABD, and having an angle equal to the given angle. b PROP. XLVI. PROB. To defcribe a fquare upon a given straight line. Let AB be the given straight line; it is required to defcribe a fquare upon AB. c d From the point A draw a AC at right angles to AB; and make AD equal to AB, and through the point D draw DE parallel to AB, and through B draw BE parallel to AD; therefore ADEB is a parallelogram; whence AB is equal to DE, and AD to BE: But BA is equal to AD; therefore the four ftraight lines BA, AD, DE, EB are equal to one another, and the parallelogram ADEB is equilateral: likewife all its angles are right angles; because the straight line AD meeting the parallels AB, DE, the angles BAD, ADE are equal to two F 2 e right a 11. 1. b 3. 1. C 31. 1. d 34.1. e 29. I. Book I. right angles; but BAD is a right angle; therefore alfo ADE is a right angle; but the oppofite angles of C d 34. 1. parallelograms are equal d; therefore each of the oppofite angles ABE, BED is a right angle; wherefore the figure ADEB is rectangular; and it has been demonstrated, that it is equilateral; it is therefore a square, and it is defcribed upon the given straight line AB. Which was to be done. D A COR. Hence every parallelogram that IN PROP. XLVII. THEOR. E B N any right angled triangle, the fquare which is defcribed upon the fide fubtending the right angle, is equal to the fquares defcribed upon the fides which contain the right angle. Let ABC be a right angled triangle having the right angle BAC; the fquare defcribed upon the fide BC is equal to the fquares defcribed upon BA, AC. a 46. I. On BC describe the fquare BDEC, and on BA, AC the fquares GB, HC; and through A draw b AL parallel to BD or CE, and join AD, FC: then, because each of the angles BAC, b 31. I. c 30. Def. BAG is a right angle, the two ftraight lines AC, AG upon the oppofite fides. of AB, make with it at the point A the adjacent angles equal to two right angles; therefore CA is in d 14. 1. the fame ftraight lined with AG; for the fame reason, AB and AH are in the fame ftraight line and because the angle DBC is equal to the angle FBA, each of them being a right angle, add to each the angle ABC, e 2. Ax. and the whole angle DBA is equal to the whole FBC; and because the two fides AB, BD are equal to the two FB, BC, each to each, and the angle DBA equal to the angle FBC; f 4. 1. therefore the bafe AD is equal f to the bafe FC, and the triangle ABD to the triangle FBC: Now the parallelogramı BL is double 41. 1. 8 of the triangle ABD, because they are upon the fame base BD, and between the fame parallels, BD, AL; and the fquare GB GB is double of the triangle FBC, because these also are upon Book I. the fame base FB, and between the fame parallels FB, GC: But the doubles of equals are equal h to one another; therefore the h 6.’Ax, parallelogram BL is equal to the fquare GB: And in the fame manner, by joining AE, BK, it may be demonftrated, that the parallelogram CL is equal to the fquare HC; therefore the whole fquare BDEC is equal to the two fquares GB, HC: and the fquare BDEC is defcribed upon the straight line BC, and the fquares GB, HC upon BA, AC: Wherefore the square upon the fide BC is equal to the fquares upon the fides BA, AC. Therefore, in any right angled triangle, &c. Q. E. D. I' PROP. XLVIII. THEOR. F the fquare described upon one of the fides of a triangle, be equal to the fquares defcribed upon the other two fides of it; the angle contained by these two fides is a right angle. If the fquare defcribed upon BC, one of the fides of the triangle ABC, be equal to the fquares upon the other fides BA, AC; the angle BAC is a right angle. b A Ꭰ From the point A draw a AD at right angles to AC, and make AD equal to BA, and join DC: Then, because DA is equal to AB, the fquare of DA is equal to the fquare of AB: To each of these add the fquare of AC; therefore the fquares of DA, AC are equal to the fquares of BA, AC: But the fquare of DC is equal to the fquares of DA, AC, because DAC is a right angle; and the square of BC, by hypothefis, is equal to the fquares of BA, AC; therefore the B fquare of DC is equal to the fquare of BC; and therefore alfo the fide DC is equal to the fide BC. And because the fide DA is equal to AB, and AC common to the two triangles DAC, BAC, the two DA, AC are equal to the two BA, AC; and the bafe DC is alfo equal to the base BC; therefore the angle DAC is equal to the angle BAC: But DAC is a right angle; therefore alfo BAC is a right angle. Therefore, if the fquare, &c. Q. E. D. a II. Is b 47. J. C 8, I THE THE ELEMENTS OF EUCLID. воок II. E DEFINITIONS. I. BOOK II. VERY right angled parallelogram is called a rectangle, and it is faid to be contained by any two of the ftraight lines which contain one of the right angles. II. A D In every parallelogram, any of the parallelograms about a dia- I' PROP. I. THEOR. 11 F G K F there be two ftraight lines, one of which is divided into any number of parts; the rectangle contained by the two ftraight lines, is equal to the rectangles contained by the undivided line, and the feveral parts of the divided line. Let b D E C a II. I. K L H b 3. I. A C 31. I. Let A and BC be two ftraight lines; and let BC be divided Book. II. into any parts in the points D, E; the rectangle contained by in the ftraight lines A, BC is equal to B the rectangle contained by A, BD, together with that contained by A, DE, and that contained by A, EC. From the point B draw a BF at right angles to BC, and make BG C equal to A; and through G draw GH parallel to BC; and through D, E, C draw c DK, EL, CH rallel to BG; then the rectangle BH is equal to the rectangles BK, DL, EH; and BH is contained by A, BC, for it is contained by GB, BC, and GB is equal to A; and BK is contained by A, BD, for it is contained by GB, BD, of which GB is equal to A; and DL is contained by A, DE, because DK, that is, BG, is equal to A; and in like manner, the rectangle EH d 34. 1. is contained by A, EC: Therefore the rectangle contained by A, BC is equal to the feveral rectangles contained by A, BD, and by A, DE; and also by A, EC. Wherefore, if there be two ftraight lines, &c. Q. E. D. d pa PROP. II. THEOR. Fa ftraight line be divided into any two parts, the rectangles contained by the whole, and each of the parts, are together equal to the fquare of the whole line. Let the ftraight line AB be divided into A any two parts in the point C; the rectangle contained by AB, BC, together with the rectangle AB, AC, fhall be equal to the fquare of AB. * Upon AB defcribe a the fquare ADEB, and through C draw b CF, parallel to AD or BE; then AE is equal to the rectangles AF, CE; and AE is the fquare of AB; D and AF is the rectangle contained by BA, св a 46. I. b 31. I. FE AC, for it is contained by DA, AC, of which AD is equal to AB; and CE is contained by AB, BC, for BE is equal to AB; therefore the rectangle contained by AB, AC, together with the rectangle *N. B. To avoid repeating the word contained too frequently, the rectangle contained by two ftraight lines AB, AC is fometimes fimply called the rectangle AB, AC. |